Is it 0 because it is being dropped from rest then stops?PhanthomJay said:
yes. And what is the change in gravitational U from book initial position to its final position?reminiscent said:
It will be 0 - (10 kg)(9.8 m/s^2)(0.25 m), correct?PhanthomJay said:
you cannot ignore the minus sign. But you are not correct here. It is given that the book is 0.25 m above the top of the spring when both are at rest in their initial position. How much is it initially from the spring when the spring will have been compressed its max distance?reminiscent said:
So are you saying that 0 - (10 kg)(9.8 m/s^2)(0.25 m) is not the correct way to calculate the change in gravitational U? How else am I supposed to solve for this when I already have 1 unknown variable?PhanthomJay said:
The unknown variable is the max spring compression, x. So if the book is .25 m above the spring before it is compressed, how high is it initially above the spring when the spring is compressed by a distance of x? You can use the unknown variable in that expression.reminiscent said:
But the book is dropped so the spring is compressed... My understanding of your explanation is that the spring compresses when the book is 0.25 above the spring, but in this problem, that is not the case.PhanthomJay said:
Remember when you calculated the change in K from initial to final position you used 0 Kinetic Energy in the initial position of the book and then 0 Kinetic Energy in the final position after the spring was compressed? Do the same for initial gravitational U and final gravitational U after the spring is compressed. When calculating gravitational U, you need to establish a reference axis for 0 gravitational U. You should take the reference 0 level for Gravitational U at the level when the spring is compressed its max distance, then calculate the change in gravitational U from start point to end point.reminiscent said:
I think I see what you are saying but I am getting confused I think... I made Ugf the reference axis for for 0 gravitational U when the spring is compressed and the book hit the spring. Ugi from my understanding is when the book is 0.25 m above the spring, so there is potential energy there. Kinetic energy in the final position is 0 since the book hit the spring so it all aligns well I think...PhanthomJay said:
This is how I set it up:PhanthomJay said:
yes, so correct your equation above for Ugireminiscent said:
well initially it is zero and its final Uelastic is just the function of its max x compression distance. Applying the conservation of energy equation, you will now get a quadratic equation to solve for x.
So everything is correct then? I got 0.25 m as the compression distance.PhanthomJay said:
No, that is not at all correct. I don't think you are paying attention. You didn't correct your Ugi equation to denote the height above the reference 0 level ...h is not equal to 0.25 m but rather h = (0.25 +x) m.reminiscent said:
I see what you are saying now. If I set the reference point 0 when it isn't compressed, would it be (0.25-x)?PhanthomJay said:
reminiscent said:
Also can I use y for spring compression instead?PhanthomJay said:
yes, you can replace x with y for all cases if you wish.reminiscent said:
Would what be that? You can set the reference point 0 for Ug at the uncompressed spring level if you want, but then the initial Ugi would be .25mg and the final Ugf would be a negative value (- mgx), and the change in U from initial to final would be Ugf - Ugi , work that out and don't let the minus signs get the best of you.reminiscent said:
Okay I used your reference point that you mentioned initially and I got 0.40 m as the max distance the spring will compress, is that correct?PhanthomJay said:
Looks good!reminiscent said:
The spring constant, also known as the force constant, can be calculated by dividing the applied force by the displacement of the spring. It is represented by the symbol k and has units of Newtons per meter (N/m).
Hooke's Law states that the force exerted by a spring is directly proportional to the amount of compression or extension of the spring. This means that as the spring is compressed, the force it exerts increases, and as it is extended, the force decreases. Therefore, Hooke's Law is essential in understanding the behavior of springs under compression.
The amount of compression in a spring is affected by several factors, including the spring constant, the applied force, and the initial length of the spring. Other factors such as the material and thickness of the spring also play a role in determining the amount of compression.
The maximum compression a spring can withstand is determined by its elastic limit. This is the point at which the spring will permanently deform or break. It is important to stay within this limit to prevent damage to the spring.
The mass of an object does not directly affect the compression of a spring. However, the force exerted by the object on the spring, as well as the amount of displacement of the spring, can vary based on the mass of the object. This can ultimately impact the amount of compression in the spring.